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Inital-value problem

  1. Mar 11, 2008 #1
    Have been looking at this problem .A few times nowI've come up with different answers.Its getting close to needing to know or wether, to just put in what i have.so ive placed a question on here to see if im missing something.

    The equations that are said to help are

    [tex]\int\f'(x)/f(x) dx = In (f(x))+c (f(x)>0)[/tex]

    dy[tex]/dx = x^4+1/x^5+5x+6 (x>-1)[/tex] is the equation to work off.

    my work so far is

    1[tex]/5 In (x^5+5)+c[/tex]
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 11, 2008 #2


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    your answer is nearly correct...you are just missing something...check back to what f(x) is and what your answer is...
  4. Mar 12, 2008 #3
    so the x^5+5x+6 does not differeniate to become x^5+5 ok

    there are a few activitys in my books that are like it.
  5. Mar 12, 2008 #4


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    [tex]\int \frac{x^4+1}{x^5+5x+6} dx[/tex]

    Let [itex]t=x^5+5x+6 \Rightarrow \frac{dt}{dx}=5(x^4+1)[/itex]


    [tex]\int \frac {x^4+1}{x^5+5x+6} dx \equiv \int \frac{1}{5} \frac{1}{t} dt[/tex]

    and [itex]\int \frac{1}{x} dx = ln(x)+C[/itex]

    [tex]\int \frac{1}{5} \frac{1}{t} dt = \frac{1}{5} lnt + C[/tex]
  6. Mar 12, 2008 #5
    that blew me a little bit there.ive sat and looked at it a little and i think i know what you have done just i would not have thought off it. i think we did one function that was worked out that way.ill have to look it up tonight.
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